it's , is it not? but we only have x on top, so we put 2a inside the integral, and 1/(2a) outside.

So you're saying that when ever we have x on top we must multiply it with the derivative of the denominator at the bottom? Then multiple the current inside integral with the derivative of the denominator?

Sorry, I asked a stupid question previously. I now see how that is formed after revising Integration by inspection. The two is actually a factor of the differentiation of the lower part of the function. We divide that by 1 or the all expression to great rid of it when it is being differentiated.